Two equal and opposite charges are placed at a certain distance apart and force of attraction between them is F. If 75% charge of one is transferred to another, then the force between the charges becomes

Options

A. `(7F)/(16)` (attraction)
B. `(F)/(16)` (attraction)
C. `(7F)/(16)` (repulsion)
D. `(F)/(16)` (repulsion)

Answer

To solve the problem step by step, we will use Coulomb's law, which states that the force \( F \) between two point charges is given by:

\[ F = k \frac{q_1 q_2}{r^2} \]

where:
- \( F \) is the force between the charges,
- \( k \) is Coulomb's constant,
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
- \( r \) is the distance between the charges.

### Step 1: Define the initial charges
Let the initial charges be:
- \( q_1 = q \) (positive charge)
- \( q_2 = -q \) (negative charge)

The initial force \( F \) between them is given by:

\[ F = k \frac{q \cdot (-q)}{r^2} = -k \frac{q^2}{r^2} \]

Since we are interested in the magnitude of the force, we can write:

\[ F = k \frac{q^2}{r^2} \]

### Step 2: Transfer 75% of charge from one to another
When 75% of charge \( q_1 \) is transferred to \( q_2 \), the amount transferred is:

\[ \text{Transferred charge} = 0.75q = \frac{3q}{4} \]

After the transfer:
- New charge \( q_1' = q - \frac{3q}{4} = \frac{q}{4} \)
- New charge \( q_2' = -q + \frac{3q}{4} = -\frac{q}{4} \)

### Step 3: Calculate the new force \( F' \)
Now we can calculate the new force \( F' \) between the modified charges \( q_1' \) and \( q_2' \):

\[ F' = k \frac{q_1' \cdot q_2'}{r^2} = k \frac{\left(\frac{q}{4}\right) \left(-\frac{q}{4}\right)}{r^2} \]

This simplifies to:

\[ F' = k \frac{-q^2/16}{r^2} = -k \frac{q^2}{16r^2} \]

Again, considering the magnitude:

\[ F' = k \frac{q^2}{16r^2} \]

### Step 4: Relate the new force to the original force
We know that the original force \( F = k \frac{q^2}{r^2} \). Therefore, we can express \( F' \) in terms of \( F \):

\[ F' = \frac{1}{16} k \frac{q^2}{r^2} = \frac{1}{16} F \]

### Step 5: Final result
Thus, the new force \( F' \) after transferring 75% of the charge is:

\[ F' = \frac{F}{16} \]

### Conclusion
The force between the charges after transferring 75% of one charge to another becomes:

\[ F' = \frac{7F}{16} \]